Pointer argument from double to double*

Question

I made a very basic program to try to better understand pointers but I cannot figure them out. I continue to get the error 'void celsius(double *)': cannot convert argument 1 from 'double' to 'double *'

#include <iostream>

using namespace std;

void celsius(double*);

int main() {

    double temp;
    double value;

    cout << "Enter temperature in C ";
    cin >> value;

    temp = celsius(value);

    cout << temp << endl;

    return 0;
}

void celsius(double *par) {
double tem;

tem = (1.8 * par + 32);
    return tem;
}

After referring to several resources I just cant figure out why it's returning the error. I know the error says it's because it can't convert a double to double* but examples I see online don't explain why.

Any help with the code or explanation be great. Thanks.

You umust take the address of `value` to pass it into a function expecting a pointer. Eg: `temp = celsius(&value);` Note: this will reveal other problems. — user4581301, Jul 13 '18 at 21:41
Just change `void celsius(double*)` to `double celsius(double)`. There is no reason the function needs a pointer. — NathanOliver, Jul 13 '18 at 21:41
Also, it sounds like you could use a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — NathanOliver, Jul 13 '18 at 21:41
`void celsuis(...` also should be changed to `double celsius(...` as you are returning a `double` — kmdreko, Jul 13 '18 at 21:41
I recommend exploring *references* before pointers. For example, declare `void celsius(double& variable)`. When passing by reference, the function knows that the parameter exists. When passing a pointer, the point may point to anywhere, including not the variable from the caller. Much safer to use references. — Thomas Matthews, Jul 13 '18 at 21:47

score 3 · Answer 1 · edited Jul 13 '18 at 21:54

3

You need to make the following changes to your program.

Change the return type of the function to double.
```
double celsius(double*);
```
When you call the function, make sure you pass a pointer to a double, not a double.
```
temp = celsius(&value);  // Use &value, not value
```
In the function, you need to use *par to dereference the pointer and use its value.
```
tem = (1.8 * (*par) + 32);  // Use *par not just par
```

Another option is to leave the return type as is but change the implementation of the function.

void celsius(double *par)
{
   *par = (1.8 * (*par) + 32);
}

and change the usage to just

celsius(&value);
cout << value << endl;

edited Jul 13 '18 at 21:54

NathanOliver

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answered Jul 13 '18 at 21:43

R Sahu

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Why not just suggest changing `void celsius(double*)` to `double celsius(double)`? It is one change instead of 3. – NathanOliver Jul 13 '18 at 21:48
@NathanOliver, I did not want to change the OP's motivation - *try to better understand pointers* – R Sahu Jul 13 '18 at 21:49
Ah, in that case `void celsius(double *par) { *par = (1.8 * (*par) + 32) }` and `celsius(value); cout << value << endl;` would showcase that more. – NathanOliver Jul 13 '18 at 21:50
@NathanOliver, that's also an option too. Thanks :) – R Sahu Jul 13 '18 at 21:53
1

Appreciate the help! This helps me understand it a lot more. Part of the issue was looking at so many different examples and not seeing what pieced together. But these examples really demonstrate it well :) – CD'A Jul 13 '18 at 22:23

score 0 · Answer 2 · answered Jul 13 '18 at 21:57

I’m sure this has been said a billion times but think of the pointer variable as just a variable.

double* is simply a type and on 32 not machines it can hold 4 Byte values. On 64 not machines there is a bigger addressing space so pointers are bigger.

Then, if double* is just a variable that can store a 4 byte address of a double then it needs the address of something...

You can take the address of other variables using ‘&’

Example:

double* ptr = &anotherVariable;

Now ‘ptr’ contains a 4 byte address and then you can access the variable contents by dereferencing the double pointer itself.

Example:

*ptr = 4.2;

However, the ‘ptr’ variable itself contains an address so the ‘anotherVariable’ that was assigned to it earlier also has the value 4.2

Now, if you were passing a value to a function and the functions parameter was ‘double* temp’. You could pass it the address of a variable or an actual pointer and inside the function you can do all the same things as above.

Thanks for this perspective on how these work, this is very helpful with my understanding. Really appreciate. — CD'A, Jul 13 '18 at 22:24

Pointer argument from double to double*

2 Answers2